Suppose the average amount of apples per tree in a large orchard is 600 pounds with a standard deviation of 70 pounds. a. Assuming the necessary conditions are satisfied, give the mean and standard deviation of the Normal model that would be used to approximate the sampling distribution of the sample mean for a sample of 40 trees. Include units in your answers.
Suppose the average amount of apples per tree in a large orchard is 600 pounds with a standard deviation of 70 pounds. a. Assuming the necessary conditions are satisfied, give the mean and standard deviation of the Normal model that would be used to approximate the sampling distribution of the sample mean for a sample of 40 trees. Include units in your answers.
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![### Problem 3: Statistical Analysis of Apple Production in Orchards
#### Context:
Suppose the average amount of apples per tree in a large orchard is 600 pounds with a standard deviation of 70 pounds.
#### Tasks:
**a. Calculation of Mean and Standard Deviation for the Sampling Distribution:**
Assuming the necessary conditions are satisfied, compute the mean and standard deviation of the Normal model to approximate the sampling distribution of the sample mean for a sample of 40 trees. Provide units in your answers.
To calculate:
- The mean of the sampling distribution \(\mu_{\bar{x}}\):
\[
\mu_{\bar{x}} = \mu = 600 \text{ pounds}
\]
- The standard deviation of the sampling distribution \(\sigma_{\bar{x}}\):
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{70}{\sqrt{40}} \approx 11.08 \text{ pounds}
\]
**b. Sketching and Labelling the Sampling Model:**
Create a sketch of the sampling model for a sample of 40 trees based on the 68-95-99.7 Rule.
- **Sketch Explanation:**
The x-axis represents the sample mean of the number of apples (in pounds), while the y-axis represents the probability density.
- The mean (μ) is 600 pounds, located centrally on the x-axis.
- Standard deviations (σ) are marked at intervals of approximately 11.08 pounds from the mean.
- The 68-95-99.7 Rule implies:
- Approximately 68% of data falls within ±1σ (588.92 to 611.08 pounds).
- Approximately 95% falls within ±2σ (577.84 to 622.16 pounds).
- Approximately 99.7% falls within ±3σ (566.76 to 633.24 pounds).
**c. Probability Calculation:**
Determine the probability that a random sample of 40 trees has an average amount of apples of 620 pounds or higher.
- Use the calculator/program of choice and specify input values.
To solve:
1. Calculate the z-score for 620 pounds:
\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{620 - 600}{11.08} \approx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Face8c1ae-f0aa-4167-9f03-8ecd729ad77b%2F1d1b7e8d-ef0c-4c38-b6a9-2199e4d4fae0%2Fqag87s4.png&w=3840&q=75)
Transcribed Image Text:### Problem 3: Statistical Analysis of Apple Production in Orchards
#### Context:
Suppose the average amount of apples per tree in a large orchard is 600 pounds with a standard deviation of 70 pounds.
#### Tasks:
**a. Calculation of Mean and Standard Deviation for the Sampling Distribution:**
Assuming the necessary conditions are satisfied, compute the mean and standard deviation of the Normal model to approximate the sampling distribution of the sample mean for a sample of 40 trees. Provide units in your answers.
To calculate:
- The mean of the sampling distribution \(\mu_{\bar{x}}\):
\[
\mu_{\bar{x}} = \mu = 600 \text{ pounds}
\]
- The standard deviation of the sampling distribution \(\sigma_{\bar{x}}\):
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{70}{\sqrt{40}} \approx 11.08 \text{ pounds}
\]
**b. Sketching and Labelling the Sampling Model:**
Create a sketch of the sampling model for a sample of 40 trees based on the 68-95-99.7 Rule.
- **Sketch Explanation:**
The x-axis represents the sample mean of the number of apples (in pounds), while the y-axis represents the probability density.
- The mean (μ) is 600 pounds, located centrally on the x-axis.
- Standard deviations (σ) are marked at intervals of approximately 11.08 pounds from the mean.
- The 68-95-99.7 Rule implies:
- Approximately 68% of data falls within ±1σ (588.92 to 611.08 pounds).
- Approximately 95% falls within ±2σ (577.84 to 622.16 pounds).
- Approximately 99.7% falls within ±3σ (566.76 to 633.24 pounds).
**c. Probability Calculation:**
Determine the probability that a random sample of 40 trees has an average amount of apples of 620 pounds or higher.
- Use the calculator/program of choice and specify input values.
To solve:
1. Calculate the z-score for 620 pounds:
\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{620 - 600}{11.08} \approx
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